A093094

[Giovanni Resta]

Klaus Brockhaus wrote:
> http://www.research.att.com/~njas/sequences/A093094
> 
> A093094: "Products into digits": start with a(1)=2, a(2)=2; repeatedly adjoin digits of product of previous two terms. 
> 2,2,4,8,3,2,2,4,6,4,8,2,4,2,4,3,2,1,6,8,8,8,1,2,6,2,6,4,8,...
> 
> A comment says that the sequence is not periodic. Perhaps this is obvious, but not to me.
> Is there a simple proof? Does it follow from a well-known theorem?

The proof is simple enough (what was not simple, for me, was to understand the
sequence definition...)

Let us assume that somewhere in the sequence there is a subsequence
of 3 adjacent 8':  ...,8,8,8,....(this is true, see above).
then we know that in the following there will be the subsequence
...,6,4,6,4.. (i.e. 8x8, 8x8) again, there will be somewhere
...,2,4,2,4,2,4,... (i.e. 6x4, 4x6, 6x4) and finally
...,8,8,8,8,8,...

Analogously, starting from 8,8,8,8 we obtain 6,4,6,4,6,4 then 2,4,2,4,2,4,2,4,2,4
and finally 8,8,8,8,8,8,8,8,8.

Generalizing, if somewhere appears a run of k>2 8's, then in some future position
will appear a run of at least 4*k-7  8's (where since k>2,  4*k-7>k).

So the sequence will contain arbitrary long runs of 8's, without
being constantly equal to 8, thus it cannot be periodic.

bye,
giovanni